Answer:
[tex]Nu_{D} = 49.047[/tex], [tex]h = 24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C}[/tex], [tex]\dot Q \approx 381.091\,W[/tex]
Explanation:
Let assume that steam has a fully developed and turbulent flow and that tube is smooth and thin-walled. The steam is heated while flowing through the tube. The Nusselt number is given by the Dittus-Boelter equation:
[tex]Nu_{D} = 0.023\cdot Re^{0.8}\cdot Pr^{0.4}[/tex]
Properties of steam at given pressure and temperature are:
[tex]c_{p} = 2.010\,\frac{kJ}{kg\cdot ^{\textdegree}C}[/tex]
[tex]k = 0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C}[/tex]
[tex]\rho = 0.5978\,\frac{kg}{m^{3}}[/tex]
[tex]\mu = 1.227\times 10^{-5}\,\frac{kg}{m\cdot s}[/tex]
The Reynolds and Prandtl numbers are, respectively:
[tex]Re = \frac{\rho\cdot v\cdot D}{\mu}[/tex]
[tex]Re = \frac{(0.5978\,\frac{kg}{m^{3}} )\cdot (6\,\frac{m}{s} )\cdot (0.05\,m)}{1.227\times 10^{-5}\,\frac{kg}{m\cdot s} }[/tex]
[tex]Re = 14616.136[/tex] (which demonstrates the reasonability of the supposition of turbulent flow)
[tex]Pr = \frac{c_{p}\cdot \mu}{k}[/tex]
[tex]Pr = \frac{(2.010\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (1.227\times 10^{-5}\,\frac{kg}{m\cdot s} )}{0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C} }[/tex]
[tex]Pr = 0.983[/tex]
The Nusselt number is:
[tex]Nu_{D} = 0.023\cdot (14616.136)^{0.8}\cdot (0.983)^{0.4}[/tex]
[tex]Nu_{D} = 49.047[/tex]
The Nusselt number has the following definition:
[tex]Nu_{D} = \frac{h\cdot D}{k}[/tex]
The heat transfer coefficient is:
[tex]h = \frac{Nu_{D}\cdot k}{D}[/tex]
[tex]h = \frac{(49.047)\cdot (0.0251\times 10^{-3}\,\frac{kW}{m\cdot ^{\textdegree}C})}{0.05\,m }[/tex]
[tex]h = 24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C}[/tex]
The convection is the dominant heat transfer mechanism, then:
[tex]\dot Q = (24.621\,\frac{W}{m^{2}\cdot ^{\textdegree}C} )\cdot [\pi\cdot (0.05\,m)\cdot (1\,m) ]\cdot (200^{\textdegree}C-100^{\textdegree}C)[/tex]
[tex]\dot Q \approx 381.091\,W[/tex]
